# Math Help - area bounded by sinx, cosx, and x-axis

1. ## area bounded by sinx, cosx, and x-axis

find the area of the region whose boundaries are given

In the first quadrant, bounded below by the x-axis and above by the curves of y=sin(x) and y=cos(x).

(A) $2-\sqrt{2}$
(B) $2+\sqrt{2}$
(C) 2
(D) $\sqrt{2}$
(E) $2\sqrt{2}$

This is what i did.
$\int_0^{\pi/4} cosx-sinx dx$

solve integral i get: $-1+\sqrt{2}$
which is not one of the multiple choice options above. I think I am not understanding what the question is asking.

2. Take a look at the graph of cos x and sin x (attached below).

The region looks like it should be split into two portions, one from 0 to pi/4, and another from pi/4 to pi/2. The function defining the top of the region is different for each one.

Can you see what the integral should be?

Edit: Oh, I see what you did. You are looking at the wrong region. Look at the one that touches the x-axis.

3. Originally Posted by drumist
Take a look at the graph of cos x and sin x (attached below).

The region looks like it should be split into two portions, one from 0 to pi/4, and another from pi/4 to pi/2. The function defining the top of the region is different for each one.

Can you see what the integral should be?

Edit: Oh, I see what you did. You are looking at the wrong region. Look at the one that touches the x-axis.
I'm still confused

4. the question does not make sense because how can it be below the x-axis and in the 1st quadrant?

5. Originally Posted by yoman360
the question does not make sense because how can it be below the x-axis and in the 1st quadrant?
No, it's bounded below by the x-axis. In other words, the x-axis is what forms the bottom-most boundary of the region. So the region is ABOVE the x-axis.

Does this make sense?

I've filled in the region it is referring to with the color green in the attached image.

6. Originally Posted by drumist
No, it's bounded below by the x-axis. In other words, the x-axis is what forms the bottom-most boundary of the region. So the region is ABOVE the x-axis.

Does this make sense?
yes. but don't know how to do it. I tried to seperate integrals one from 0 to $\pi/4$ and another from $\pi/4$ to $\pi/2$ (as you posted above) but i still get incorrect answer.

7. OK. The first half is bounded above by sin x, and the second half is bounded above by cos x.

$\int_0^{\pi / 4} \sin x \, dx + \int_{\pi / 4}^{\pi / 2} \cos x \, dx$

$= [-\cos x]_0^{\pi / 4} + [\sin x]_{\pi / 4}^{\pi / 2}$

$= -\cos (\pi/4) -(- \cos 0) + \sin (\pi / 2) - \sin(\pi / 4)$

You can then reduce this down to one of the answer choices that you listed.

Let me know if you can't follow what I am doing here.

8. Originally Posted by drumist
OK. The first half is bounded above by sin x, and the second half is bounded above by cos x.

$\int_0^{\pi / 4} \sin x \, dx + \int_{\pi / 4}^{\pi / 2} \cos x \, dx$

$= [-\cos x]_0^{\pi / 4} + [\sin x]_{\pi / 4}^{\pi / 2}$

$= -\cos (\pi/4) -(- \cos 0) + \sin (\pi / 2) - \sin(\pi / 4)$

You can then reduce this down to one of the answer choices that you listed.

Let me know if you can't follow what I am doing here.
thanks